10—Partial Differential Equations 259whereknmis thek 3 of the preceding equation. What happened to the other term inz, the one with
the positive exponent? Did I say that I’m looking for solutions in the domainz > 0?
Atz= 0this must match the boundary conditions stated, and as before, the orthogonality of the
sines on the two domains allows you to determine the coefficients. You simply have to do two integrals
instead of one. See problem10.19.
V(x,y,z >0) =
1
2
V 0 +
8 V 0
π^2
∑∞
oddn∑∞
oddm1
nm
sin(nπx
a
)
sin(mπy
b
)
e−knmz (10.48)
10.7 Cylindrical Coordinates
Rectangular coordinates are not always the right choice. Cylindrical, spherical, and other choices are
often needed. For cylindrical coordinates, the gradient and divergence are, from Eqs. (9.24) and (9.15)
∇V =rˆ
∂V
∂r
+φˆ
1
r
∂V
∂φ
+ˆz
∂V
∂z
and ∇.~v=
1
r
∂(rvr)
∂r
+
1
r
∂vφ
∂φ
+
∂vz
∂z
Combine these to get the Laplacian in cylindrical coordinates.
∇.∇V=∇^2 V=
1
r
∂
∂r
(
r
∂V
∂r
)
+
1
r^2
∂^2 V
∂φ^2
+
∂^2 V
∂z^2
(10.49)
For electrostatics, the equation remains∇^2 V = 0, and you can approach it the same way as before,
using separation of variables. I’ll start with the special case for which everything is independent ofz.
Assume a solution of the formV=f(r)g(φ), then
∇^2 V =g
1
r
∂
∂r
(
r
∂f
∂r
)
+f
1
r^2
∂^2 g
∂φ^2
= 0
Multiply this byr^2 and divide byf(r)g(φ)to get
r
f
∂
∂r
(
r
∂f
∂r
)
+
1
g
∂^2 g
∂φ^2
= 0
This is separated. The first terms depends onralone, and the second term onφalone. For this to
hold the terms must be constants.